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Schach [20]
3 years ago
9

PL-1) A spring that hangs vertically is 25 cm long when no weight is attached to its lower end. Steve adds 250 g of mass to the

end of the spring, which stretches to a new length of 37 cm. What is the spring constant, k, in N/m?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

Answer:

20.42 N/m

Explanation:

From hook's law,

F = ke ......................... Equation 1

Where F = Force applied to the spring., k = spring constant, e = extension.

Make k the subject of the equation,

k = F/e ................. Equation 2

Note: The force on the spring is equal to the weight of the mass hung on it.

F = W = mg.

k = mg/e................ Equation 3

Given: m = 250 g = 0.25 kg, e = 37-25 = 12 cm = 0.12 m.

Constant: g = 9.8 m/s²

Substitute into equation 3

k = (0.25×9.8)/0.12

k = 20.42 N/m.

Hence the spring constant = 20.42 N/m

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Answer:

wavelength decreases and frequency increase

Explanation:

the higher the wavelength the smaller the frequency , the smaller the wavelength the higher the frequency

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3 years ago
An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an
Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of \gamma

Using formula of relativistic energy

E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

Put the value into the formula

1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}

Here, \gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

\gamma-1=0.01953

\gamma=0.01953+1

\gamma=1.01953

We need to calculate the length

Using formula of length

L'=\dfrac{L}{\gamma}

Put the value into the formula

L'=\dfrac{4}{1.01953}

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Hence, The length of the tube is 3.92 m.

8 0
3 years ago
A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Zolol [24]
     Considering the unknown resistence as R and using the Ohm's First Law, we have:

i= \frac{V}{R_{eq}}  \\ 0.5= \frac{12}{R+10}  \\ R+10=24 \\ R=14-Ohm
 
     The equivalent resistence is given by the resistor series with the lamp resistence.

R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}

If you notice any mistake in my english, please let me know, because i am not native.

6 0
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An incompressible fluid (water) is flowing through a pipe of diameter 20 cm with
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Answer:

115 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Assuming no elevation change, h₁ = h₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Plugging in values:

(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²

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3 0
3 years ago
How are the wavelength and frequency of electromagnetic waves related to their energy
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As wavelength decreases, frequency increases, but as frequency decreases, wavelength increases...Vice-Versa
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